149557
The radiated power of a body at $400 \mathrm{~K}$ is 1000 $W$. If the temperature is raised to $800 \mathrm{~K}$, what would be the radiated power of the body?
1 $12000 \mathrm{~W}$
2 $15000 \mathrm{~W}$
3 $16000 \mathrm{~W}$
4 $18000 \mathrm{~W}$
Explanation:
C Given $\mathrm{T}_{1}=400 \mathrm{~K}, \mathrm{~T}_{2}=800 \mathrm{~K}$ and $\mathrm{E}_{1}=$ $1000 \mathrm{~W}$ We know from the Stefan's law $\mathrm{E} \propto \mathrm{T}^{4}$ $\text { Hence, } \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{1000}{\mathrm{E}_{2}}=\frac{(400)^{4}}{(800)^{4}} \Rightarrow \frac{1000}{\mathrm{E}_{2}}=\frac{1}{(2)^{4}}$ $\mathrm{E}_{2}=16000 \mathrm{~W}$
TS EAMCET 06.08.2021
Heat Transfer
149558
Two spheres of same material and radii $5 \mathrm{~m}$ and $2 \mathrm{~m}$ are at temperature $2000 \mathrm{~K}$ and $2500 \mathrm{~K}$ respectively. The ratio of energies radiated by them per second is
1 $64: 25$
2 $36: 75$
3 $128: 625$
4 $16: 125$
Explanation:
A Given- $\text { } \mathrm{R}_{1}=5 \mathrm{~m}$ $\mathrm{R}_{2}=2 \mathrm{~m}$ $\mathrm{T}_{1}=2000 \mathrm{~K}, \mathrm{~T}_{2}=2500 \mathrm{~K}$ By using Stefan's law, energy radiated per second. $\mathrm{E}=\sigma \mathrm{A} \in \mathrm{T}^{4}$ $\sigma=$ Stefan's constant $=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}$ $\mathrm{A}=$ Area, and $\in=$ emissivity $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} \propto \frac{\mathrm{A}_{1} \mathrm{~T}_{1}^{4}}{\mathrm{~A}_{2} \mathrm{~T}_{2}^{4}} \quad[\because \in$ is same for both body $]$ $\frac{E_{1}}{E_{2}}=\frac{4 \pi R_{1}^{2}}{4 \pi R_{2}^{2}} \times\left(\frac{T_{1}}{T_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2} \times\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{5}{2}\right)^{2} \times\left(\frac{2000}{2500}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{25}{4} \times\left(\frac{4}{5}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{25}{4} \times \frac{4^{4}}{5^{2} \times 5^{2}}=\frac{64}{25}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=64: 25$
TS EAMCET 04.08.2021
Heat Transfer
149559
The temperature of the sun can be found out by using
1 Wien's displacement law
2 Kepler's law of motion
3 Stefan's Boltzmann law
4 Planck's law
Explanation:
C According to Stefan's law the radiant energy emitted by a perfectly black body per unit area per second is directly proportional to the fourth power of its absolute temperature i.e. $\mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\mathrm{E}=$ amount of heat energy radiated/sec area. $\sigma=\text { Stefan's Boltzmann constant. }$ - Wien's law is useful for estimating the surface temperature of celestial bodies like the moon, sun and other stars. - Kepler's law described the orbits of planets around the sun or stars. - Planck's law is to explain the spectral-energy distribution of direction radiation emitted by black body that absorbs all radiant energy falling upon it, reaches some equilibrium temperature and then reemits.
JIPMER-2011
Heat Transfer
149560
The rate of radiation energy from high temperature black body at $\mathrm{T} \mathrm{K}$ is $\mathrm{EW} / \mathrm{m}^{2}$ What will be the rate of radiation, if temperature decreases to $\left(\frac{2 T}{3}\right) K$ ?
1 $\frac{8 \mathrm{E}}{27}$
2 $\frac{16 \mathrm{E}}{27}$
3 $\frac{16 \mathrm{E}}{81}$
4 $\frac{32 \mathrm{E}}{81}$
Explanation:
C As by Stefan's formula $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\mathrm{E} \propto \mathrm{T}^{4}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{T}_{2}=\frac{2 \mathrm{~T}}{3}$ $\mathrm{T}_{1}=\mathrm{T}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{T}_{2}^{4}}{\mathrm{~T}_{1}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\left(\frac{2 \mathrm{~T}}{3}\right)^{4}}{(\mathrm{~T})^{4}}=\frac{\frac{16}{81} \mathrm{~T}^{4}}{\mathrm{~T}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}}=\frac{16}{81}$ $\mathrm{E}_{2}=\frac{16}{81} \mathrm{E}$
149557
The radiated power of a body at $400 \mathrm{~K}$ is 1000 $W$. If the temperature is raised to $800 \mathrm{~K}$, what would be the radiated power of the body?
1 $12000 \mathrm{~W}$
2 $15000 \mathrm{~W}$
3 $16000 \mathrm{~W}$
4 $18000 \mathrm{~W}$
Explanation:
C Given $\mathrm{T}_{1}=400 \mathrm{~K}, \mathrm{~T}_{2}=800 \mathrm{~K}$ and $\mathrm{E}_{1}=$ $1000 \mathrm{~W}$ We know from the Stefan's law $\mathrm{E} \propto \mathrm{T}^{4}$ $\text { Hence, } \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{1000}{\mathrm{E}_{2}}=\frac{(400)^{4}}{(800)^{4}} \Rightarrow \frac{1000}{\mathrm{E}_{2}}=\frac{1}{(2)^{4}}$ $\mathrm{E}_{2}=16000 \mathrm{~W}$
TS EAMCET 06.08.2021
Heat Transfer
149558
Two spheres of same material and radii $5 \mathrm{~m}$ and $2 \mathrm{~m}$ are at temperature $2000 \mathrm{~K}$ and $2500 \mathrm{~K}$ respectively. The ratio of energies radiated by them per second is
1 $64: 25$
2 $36: 75$
3 $128: 625$
4 $16: 125$
Explanation:
A Given- $\text { } \mathrm{R}_{1}=5 \mathrm{~m}$ $\mathrm{R}_{2}=2 \mathrm{~m}$ $\mathrm{T}_{1}=2000 \mathrm{~K}, \mathrm{~T}_{2}=2500 \mathrm{~K}$ By using Stefan's law, energy radiated per second. $\mathrm{E}=\sigma \mathrm{A} \in \mathrm{T}^{4}$ $\sigma=$ Stefan's constant $=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}$ $\mathrm{A}=$ Area, and $\in=$ emissivity $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} \propto \frac{\mathrm{A}_{1} \mathrm{~T}_{1}^{4}}{\mathrm{~A}_{2} \mathrm{~T}_{2}^{4}} \quad[\because \in$ is same for both body $]$ $\frac{E_{1}}{E_{2}}=\frac{4 \pi R_{1}^{2}}{4 \pi R_{2}^{2}} \times\left(\frac{T_{1}}{T_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2} \times\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{5}{2}\right)^{2} \times\left(\frac{2000}{2500}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{25}{4} \times\left(\frac{4}{5}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{25}{4} \times \frac{4^{4}}{5^{2} \times 5^{2}}=\frac{64}{25}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=64: 25$
TS EAMCET 04.08.2021
Heat Transfer
149559
The temperature of the sun can be found out by using
1 Wien's displacement law
2 Kepler's law of motion
3 Stefan's Boltzmann law
4 Planck's law
Explanation:
C According to Stefan's law the radiant energy emitted by a perfectly black body per unit area per second is directly proportional to the fourth power of its absolute temperature i.e. $\mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\mathrm{E}=$ amount of heat energy radiated/sec area. $\sigma=\text { Stefan's Boltzmann constant. }$ - Wien's law is useful for estimating the surface temperature of celestial bodies like the moon, sun and other stars. - Kepler's law described the orbits of planets around the sun or stars. - Planck's law is to explain the spectral-energy distribution of direction radiation emitted by black body that absorbs all radiant energy falling upon it, reaches some equilibrium temperature and then reemits.
JIPMER-2011
Heat Transfer
149560
The rate of radiation energy from high temperature black body at $\mathrm{T} \mathrm{K}$ is $\mathrm{EW} / \mathrm{m}^{2}$ What will be the rate of radiation, if temperature decreases to $\left(\frac{2 T}{3}\right) K$ ?
1 $\frac{8 \mathrm{E}}{27}$
2 $\frac{16 \mathrm{E}}{27}$
3 $\frac{16 \mathrm{E}}{81}$
4 $\frac{32 \mathrm{E}}{81}$
Explanation:
C As by Stefan's formula $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\mathrm{E} \propto \mathrm{T}^{4}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{T}_{2}=\frac{2 \mathrm{~T}}{3}$ $\mathrm{T}_{1}=\mathrm{T}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{T}_{2}^{4}}{\mathrm{~T}_{1}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\left(\frac{2 \mathrm{~T}}{3}\right)^{4}}{(\mathrm{~T})^{4}}=\frac{\frac{16}{81} \mathrm{~T}^{4}}{\mathrm{~T}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}}=\frac{16}{81}$ $\mathrm{E}_{2}=\frac{16}{81} \mathrm{E}$
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Heat Transfer
149557
The radiated power of a body at $400 \mathrm{~K}$ is 1000 $W$. If the temperature is raised to $800 \mathrm{~K}$, what would be the radiated power of the body?
1 $12000 \mathrm{~W}$
2 $15000 \mathrm{~W}$
3 $16000 \mathrm{~W}$
4 $18000 \mathrm{~W}$
Explanation:
C Given $\mathrm{T}_{1}=400 \mathrm{~K}, \mathrm{~T}_{2}=800 \mathrm{~K}$ and $\mathrm{E}_{1}=$ $1000 \mathrm{~W}$ We know from the Stefan's law $\mathrm{E} \propto \mathrm{T}^{4}$ $\text { Hence, } \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{1000}{\mathrm{E}_{2}}=\frac{(400)^{4}}{(800)^{4}} \Rightarrow \frac{1000}{\mathrm{E}_{2}}=\frac{1}{(2)^{4}}$ $\mathrm{E}_{2}=16000 \mathrm{~W}$
TS EAMCET 06.08.2021
Heat Transfer
149558
Two spheres of same material and radii $5 \mathrm{~m}$ and $2 \mathrm{~m}$ are at temperature $2000 \mathrm{~K}$ and $2500 \mathrm{~K}$ respectively. The ratio of energies radiated by them per second is
1 $64: 25$
2 $36: 75$
3 $128: 625$
4 $16: 125$
Explanation:
A Given- $\text { } \mathrm{R}_{1}=5 \mathrm{~m}$ $\mathrm{R}_{2}=2 \mathrm{~m}$ $\mathrm{T}_{1}=2000 \mathrm{~K}, \mathrm{~T}_{2}=2500 \mathrm{~K}$ By using Stefan's law, energy radiated per second. $\mathrm{E}=\sigma \mathrm{A} \in \mathrm{T}^{4}$ $\sigma=$ Stefan's constant $=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}$ $\mathrm{A}=$ Area, and $\in=$ emissivity $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} \propto \frac{\mathrm{A}_{1} \mathrm{~T}_{1}^{4}}{\mathrm{~A}_{2} \mathrm{~T}_{2}^{4}} \quad[\because \in$ is same for both body $]$ $\frac{E_{1}}{E_{2}}=\frac{4 \pi R_{1}^{2}}{4 \pi R_{2}^{2}} \times\left(\frac{T_{1}}{T_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2} \times\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{5}{2}\right)^{2} \times\left(\frac{2000}{2500}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{25}{4} \times\left(\frac{4}{5}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{25}{4} \times \frac{4^{4}}{5^{2} \times 5^{2}}=\frac{64}{25}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=64: 25$
TS EAMCET 04.08.2021
Heat Transfer
149559
The temperature of the sun can be found out by using
1 Wien's displacement law
2 Kepler's law of motion
3 Stefan's Boltzmann law
4 Planck's law
Explanation:
C According to Stefan's law the radiant energy emitted by a perfectly black body per unit area per second is directly proportional to the fourth power of its absolute temperature i.e. $\mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\mathrm{E}=$ amount of heat energy radiated/sec area. $\sigma=\text { Stefan's Boltzmann constant. }$ - Wien's law is useful for estimating the surface temperature of celestial bodies like the moon, sun and other stars. - Kepler's law described the orbits of planets around the sun or stars. - Planck's law is to explain the spectral-energy distribution of direction radiation emitted by black body that absorbs all radiant energy falling upon it, reaches some equilibrium temperature and then reemits.
JIPMER-2011
Heat Transfer
149560
The rate of radiation energy from high temperature black body at $\mathrm{T} \mathrm{K}$ is $\mathrm{EW} / \mathrm{m}^{2}$ What will be the rate of radiation, if temperature decreases to $\left(\frac{2 T}{3}\right) K$ ?
1 $\frac{8 \mathrm{E}}{27}$
2 $\frac{16 \mathrm{E}}{27}$
3 $\frac{16 \mathrm{E}}{81}$
4 $\frac{32 \mathrm{E}}{81}$
Explanation:
C As by Stefan's formula $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\mathrm{E} \propto \mathrm{T}^{4}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{T}_{2}=\frac{2 \mathrm{~T}}{3}$ $\mathrm{T}_{1}=\mathrm{T}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{T}_{2}^{4}}{\mathrm{~T}_{1}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\left(\frac{2 \mathrm{~T}}{3}\right)^{4}}{(\mathrm{~T})^{4}}=\frac{\frac{16}{81} \mathrm{~T}^{4}}{\mathrm{~T}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}}=\frac{16}{81}$ $\mathrm{E}_{2}=\frac{16}{81} \mathrm{E}$
149557
The radiated power of a body at $400 \mathrm{~K}$ is 1000 $W$. If the temperature is raised to $800 \mathrm{~K}$, what would be the radiated power of the body?
1 $12000 \mathrm{~W}$
2 $15000 \mathrm{~W}$
3 $16000 \mathrm{~W}$
4 $18000 \mathrm{~W}$
Explanation:
C Given $\mathrm{T}_{1}=400 \mathrm{~K}, \mathrm{~T}_{2}=800 \mathrm{~K}$ and $\mathrm{E}_{1}=$ $1000 \mathrm{~W}$ We know from the Stefan's law $\mathrm{E} \propto \mathrm{T}^{4}$ $\text { Hence, } \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{1000}{\mathrm{E}_{2}}=\frac{(400)^{4}}{(800)^{4}} \Rightarrow \frac{1000}{\mathrm{E}_{2}}=\frac{1}{(2)^{4}}$ $\mathrm{E}_{2}=16000 \mathrm{~W}$
TS EAMCET 06.08.2021
Heat Transfer
149558
Two spheres of same material and radii $5 \mathrm{~m}$ and $2 \mathrm{~m}$ are at temperature $2000 \mathrm{~K}$ and $2500 \mathrm{~K}$ respectively. The ratio of energies radiated by them per second is
1 $64: 25$
2 $36: 75$
3 $128: 625$
4 $16: 125$
Explanation:
A Given- $\text { } \mathrm{R}_{1}=5 \mathrm{~m}$ $\mathrm{R}_{2}=2 \mathrm{~m}$ $\mathrm{T}_{1}=2000 \mathrm{~K}, \mathrm{~T}_{2}=2500 \mathrm{~K}$ By using Stefan's law, energy radiated per second. $\mathrm{E}=\sigma \mathrm{A} \in \mathrm{T}^{4}$ $\sigma=$ Stefan's constant $=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}$ $\mathrm{A}=$ Area, and $\in=$ emissivity $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} \propto \frac{\mathrm{A}_{1} \mathrm{~T}_{1}^{4}}{\mathrm{~A}_{2} \mathrm{~T}_{2}^{4}} \quad[\because \in$ is same for both body $]$ $\frac{E_{1}}{E_{2}}=\frac{4 \pi R_{1}^{2}}{4 \pi R_{2}^{2}} \times\left(\frac{T_{1}}{T_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2} \times\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{5}{2}\right)^{2} \times\left(\frac{2000}{2500}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{25}{4} \times\left(\frac{4}{5}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{25}{4} \times \frac{4^{4}}{5^{2} \times 5^{2}}=\frac{64}{25}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=64: 25$
TS EAMCET 04.08.2021
Heat Transfer
149559
The temperature of the sun can be found out by using
1 Wien's displacement law
2 Kepler's law of motion
3 Stefan's Boltzmann law
4 Planck's law
Explanation:
C According to Stefan's law the radiant energy emitted by a perfectly black body per unit area per second is directly proportional to the fourth power of its absolute temperature i.e. $\mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\mathrm{E}=$ amount of heat energy radiated/sec area. $\sigma=\text { Stefan's Boltzmann constant. }$ - Wien's law is useful for estimating the surface temperature of celestial bodies like the moon, sun and other stars. - Kepler's law described the orbits of planets around the sun or stars. - Planck's law is to explain the spectral-energy distribution of direction radiation emitted by black body that absorbs all radiant energy falling upon it, reaches some equilibrium temperature and then reemits.
JIPMER-2011
Heat Transfer
149560
The rate of radiation energy from high temperature black body at $\mathrm{T} \mathrm{K}$ is $\mathrm{EW} / \mathrm{m}^{2}$ What will be the rate of radiation, if temperature decreases to $\left(\frac{2 T}{3}\right) K$ ?
1 $\frac{8 \mathrm{E}}{27}$
2 $\frac{16 \mathrm{E}}{27}$
3 $\frac{16 \mathrm{E}}{81}$
4 $\frac{32 \mathrm{E}}{81}$
Explanation:
C As by Stefan's formula $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\mathrm{E} \propto \mathrm{T}^{4}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{T}_{2}=\frac{2 \mathrm{~T}}{3}$ $\mathrm{T}_{1}=\mathrm{T}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{T}_{2}^{4}}{\mathrm{~T}_{1}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\left(\frac{2 \mathrm{~T}}{3}\right)^{4}}{(\mathrm{~T})^{4}}=\frac{\frac{16}{81} \mathrm{~T}^{4}}{\mathrm{~T}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}}=\frac{16}{81}$ $\mathrm{E}_{2}=\frac{16}{81} \mathrm{E}$